Question

Simplify the given expression below:

4 divided by the quantity of 3 minus 2i

Answer 1

\[\frac{ 4 }{ 3-2i }\]

Answer 2

@The_One_And_Only wouldn't I multiply the numerator and denominator by -1?

Answer 3

meant @TheSmartOne

Answer 4

I would think that you would need to multiply the numerator and denominator by the conjugate of 3 - 2i so that way there is no i in the denominator.

Answer 5

First fiind the complex conjugate of 3-2i and multiply both nummerator and denominator by it:\[\frac{4}{3-2i} \times\frac{conj}{conj}\]

Answer 6

did that and got \[\frac{ 12-8i }{ 9+2i }\]

Answer 7

\(\sf (a-bi)(a+bi) = a^2 + b^2\)

Try again.

Answer 8

The i should be complete gone because i^2 = -1

and

(a-b)(a+b) = a^2 - b^2

and then the i^2 is -1

so it makes it a^2 + b^2 as you can see in the formula I gave you

Answer 9

Not understanding the options give a 13 at the denominator not sure how I would get 13 for it keep getting 7 or 11

Answer 10

\(\sf\Large \frac{4}{3-2i} \times \frac{3+2i}{3+2i} = \frac{4(3+2i)}{(3-2i)(3+2i)}\)

Now,

\(\sf (a-bi)(a+bi) =a^2 + b^2\\\\(3-2i)(3+2i) = ??\)

Answer 11

9+6i-6i-4i
9+4i

Answer 12

would I be canceled out since (a+b)^2

Answer 13

or put it up with the numerator

Answer 14

\(\ 2i \times 2i = 4i^2\)

and \(\sf i^2 = -1\)

so \(\sf 4i^2 = -4\)

Answer 15

9 - 4i^2 = 9 - (-4) = 9 + 4 = 13

so now we have:

\(\sf\Large \frac{4}{3-2i} \times \frac{3+2i}{3+2i} = \frac{4(3+2i)}{(3-2i)(3+2i)}=\frac{4(3+2i)}{13}\)

Answer 16

so\[\frac{ 12+8i }{ 13 }\]

Answer 17

thank yall for the help

Answer 18

Anytime!